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a(n) = (A005704(n)-A006047(n))/3.
3

%I #20 Dec 18 2023 08:26:17

%S 0,0,0,1,1,1,3,3,3,7,8,9,12,13,14,19,20,21,30,33,36,42,45,48,57,60,63,

%T 79,86,93,103,111,119,132,141,150,168,180,192,209,222,235,257,271,285,

%U 316,335,354,380,400,420,453,474,495,543,573,603,639,672,705,747

%N a(n) = (A005704(n)-A006047(n))/3.

%H G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.122.9.880">Characterizing the number of m-ary partitions modulo m</a>, The American Mathematical Monthly, Vol. 122, No. 9 (November 2015), pp. 880-885.

%H G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, <a href="https://georgeandrews1.github.io/pdf/300.pdf">Characterizing the number of m-ary partitions modulo m</a>.

%H Tom Edgar, <a href="http://arxiv.org/abs/1603.00085">The distribution of the number of parts of m-ary partitions modulo m.</a>, arXiv:1603.00085 [math.CO], 2016.

%F Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/3)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*3^i is the ternary representation of n. Then a(n) = (1/3)*(b(n) - c(n)).

%t b[n_] := b[n] = If[n <= 2, n+1, b[n-1] + b[Floor[n/3]]];

%t c = Nest[Join[#, 2#, 3#]&, {1}, 4];

%t a[n_] := (b[n] - c[[n+1]])/3;

%t Table[a[n], {n, 0, 60}] (* _Jean-François Alcover_, Dec 12 2018 *)

%o (Sage)

%o def b(n):

%o A=[1]

%o for i in [1..n]:

%o A.append(A[i-1] + A[floor(i/3)])

%o return A[n]

%o [(b(n)-prod(x+1 for x in n.digits(3)))/3 for n in [0..60]]

%Y Cf. A005704, A006047, A268128.

%K nonn

%O 0,7

%A _Tom Edgar_, Jan 26 2016